when you do web2py.py -S .....
you're starting a process that presents a shell with the web2py environment: it doesn't serve any web pages in fact, "-S" stands for "shell". if you want the queue to be executed, it's correct to start it with -S but you must start ANOTHER process for serving the web part, i.e. web2py.py -a '<recycle>' -i 127.0.0.1 -p 8000 On Saturday, April 16, 2016 at 6:53:35 AM UTC+2, Smruti Mohanty wrote: > > I have an application written in web2py framework.I have a task queue > which polls for every 60 seconds. The Task queue(tqueue.py) contains a > program which inserts into a database. > > > But when I do: > > > python web2py.py -S app1 -i 127.0.0.1 -p 9001 -M -R > applications/app1/private/tqueue.py > > > The cmd prompts shows the following information: > > > web2py Web FrameworkCreated by Massimo Di Pierro, Copyright 2007-2016Version > 2.14.3-stable+timestamp.2016.03.26.23.02.02Database drivers available: > psycopg2, pymysql, imaplib, sqlite3, pg8000, pyodbc > > > But when i try to open the webpage: > > http://127.0.0.1:9001/app1/default/index > > > It doesnot open up and says connection refused. I could see the output > from tqueue.py script though. I have some print statement inside the > tqueue.py script which prints to the command prompt. > > > When I try to do only: > > > python web2py.py > > > I can reach the administrative interface as well as open the Homepage at > http://127.0.0.1:9001/app1/default/index > > But my task queue becomes inactive. > > > Kindly let me know if i am doing anything wrong. > > > > > -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
